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11.2 (1)

11.2 (1) - set the KE of each gas equal to each other like...

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Key: 2. At the same temperature, water vapor molecules will diffuse ________ times _________ than carbon dioxide molecules. a. 6.4, faster b. 1.6, faster c. 12.0, faster d. 6.4, slower e. 1.6, slower f. 12.0 slower This problem asks how much faster one gas will diffuse than another when both gases are at the same temperature. In other words, it is asking for the ratio of the velocity of the water vapor over the velocity of the carbon dioxide gas. This could be written as follows: v H20 /v CO2 = ? In order to solve this problem, you need to first understand that the average kinetic energy (KE) of any two gases is equal when the temperatures of those two gases are equal. We can therefore
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Unformatted text preview: set the KE of each gas equal to each other like so: KE H2O = KE CO2 Hopefully you remember that KE = 1/2mv 2 . We can now adjust the formula to 1/2m H2O (v H2O ) 2 = 1/2m CO2 (v CO2 ) 2 Now you can use algebra to rearrange the formula so it equals the ratio we are looking for. Our formula now looks like this: v H2O /v CO2 = √(m CO2 /m H2O ). Substituting the molar masses of CO 2 and H 2 O into the equation, we get: v H2O /v CO2 = √44/18 = 1.6/1. In other words, v H2O = 1.6 when v CO2 =1. That means water diffuses 1.6 times faster than carbon dioxide. ....
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